math110s-hw8sol

math110s-hw8sol - MATH 110: LINEAR ALGEBRA SPRING 2007/08...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MATH 110: LINEAR ALGEBRA SPRING 2007/08 PROBLEM SET 8 SOLUTIONS 1. Let V be a vector space over F and S , T End( V ). (a) Show that I - S T is injective iff I - T S is injective. Solution. The trick is to observe that S ( I - T S ) = ( I - S T ) S . Suppose I -S T is injective. Let v ker( I -T S ). Then ( I -T S )( v ) = V . Hence ( I - S T )( S ( v )) = S (( I - T S )( v )) = S ( V ) = V and since I - S T is injective, we deduce that S ( v ) = V . So T S ( v ) = T ( V ) = V ; and so v = v- V = v- T S ( v ) = ( I - T S )( v ) = V where the last equality follows from our choice of v ker( I-T S ). Hence ker( I-T S ) = { V } and we deduce that I - T S is injective by Theorem 4.12 . (b) T is called nilpotent if T n = O for some n N . Show that if T is nilpotent, then I - T is bijective. What is ( I - T )- 1 ? Solution. Let S = I + T + T 2 + + T n- 1 (by the geometric series heuristic in lecture). Then ( I - T ) S = ( I + T + + T n- 1 )- ( T + T 2 + + T n ) = I - T n = I - O = I and likewise for S ( I - T ). Hence ( I - T )- 1 = S and I - T is bijective. 2. Let V be a vector space over R and T End( V ) be an involution, ie. T 2 = I . Define V + := { v V | T ( v ) = v } and V- := { v V | T ( v ) =- v } . (a) Show that V + and V- are subspaces of V . Solution. Let 1 , 2 R and v 1 , v 2 V + . Then T ( v 1 ) = v 1 , T ( v 2 ) = v 2 , and T ( 1 v 1 + 2 v 2 ) = 1 T ( v 1 ) + 2 T ( v 2 ) = 1 v 1 + 2 v 2 . Hence 1 v 1 + 2 v 2 V + . Likewise for V- . Date : May 7, 2008 (Version 1.0). 1 (b) Show that V + V- = V. Solution. Note that v = 1 2 ( I + T )( v ) + 1 2 ( I - T )( v ) . (2.1) Since T 2 = I , T ( 1 2 ( I + T )( v ) ) = T ( 1 2 v + 1 2 T ( v ) ) = 1 2 T ( v ) + 1 2 T 2 ( v ) = 1 2 T ( v ) + 1 2 I ( v ) = 1 2 ( I + T )( v ) and likewise, T ( 1 2 ( I - T )( v ) ) = T ( 1 2 v- 1 2 T ( v ) ) = 1 2 T ( v )- 1 2 T 2 ( v ) = 1 2 T ( v )- 1 2 I ( v ) =- 1 2 ( I - T )( v ) ....
View Full Document

Page1 / 5

math110s-hw8sol - MATH 110: LINEAR ALGEBRA SPRING 2007/08...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online